使用时间序列参数的Python优化Pyomo

Question

我正在优化一种能源系统，该能源系统从两种天然气和氢来提供能源为一周内（每小时时间步长）内需求变化的最终用户。

我使用PYOMO和IPOPT求解器对正常工作的系统进行建模。我在两个单独的列中有两个在CSV文件中的入口参数（需求和氢）的列表（正如我所说，有一个小时的时间步长）。

我想为每个时间步骤优化系统，并生成OF并计算每个时间步长的变量（所有行的每小时）。错误是

索引'（1，'ng'）'对于索引组件'x'

无效

我认为将函数用于每个变量和约束“ def”，但我无法成功获得结果。

如果您可以看一下代码，并告诉我如何重复每个时间步骤的优化，我将不胜感激。

    import pyomo.environ as pyo
    from pyomo.opt import SolverFactory
    from pyomo.environ import *
    import pandas as pd
    import matplotlib as mp
    import matplotlib.pyplot as plt 
    from pyomo.dae import *
    import numpy as np
    
    data=pd.read_csv('H2_Demand_Week.csv')
    
     
    date_numbers = data['Time']
    
    y = data['Demand_MWh'] 
    y2 = data['H2_Curtailed_10MWh']
    
    m = pyo.ConcreteModel()
    Nd=len(data)
    
    
     
    cost_NG = 0.066 #the NG price is 0.066 €/kWh in Ireland in March 2022
    cost_CW = 25 #curtailed wind cost is 25 €/MWh
    cost_Storage = 470 #€/MWh
    cost_H2_out = 0.044 #€/nm3
    cost_H2_in = 0.5 #€/nm3
    WI_NG = 48  #the natural gas Wobbe Index MJ/m3
    WI_H2 = 41 #the hydrogen Wobbe Index MJ/m3
    
   
    
   
H2_perc=0.05

yd = data['Demand_MWh'] 
yh = data['H2_Curtailed_10MWh']


T = range(len(data))


data.replace([np.nan], 0, inplace=True)

#sets and paramenters
#m.setInv = pyo.Set(initialize=['NG','H','S'])

m.NG = pyo.Var(T, bounds=(0,None), initialize=1)

m.H = pyo.Var(T, bounds=(0,None), initialize=1)

m.S = pyo.Var(T, bounds=(0,None), initialize=1)

#constraints
#m.C1 = pyo.Constraint(expr = pyo.summation(m.X) == Demand)
for i in range(1,Nd):
    m.C1 = pyo.Constraint(expr = m.NG[i] + m.H[i] ==yd[i])
    

for i in range(1,Nd):    
    m.C2 = pyo.Constraint(expr = m.H[i] + m.S[i] == yh[i])
    

for i in range(1,Nd):    
    m.C3 = pyo.Constraint(expr = m.H[i]/(m.NG[i] + m.H[i])<= H2_perc) #hydrogen concentration


#objective function
#m.obj = pyo.Objective(expr = m.X['NG']*0.066+m.X['H']*0.5+m.X['S']*cost_Storage, sense=pyo.minimize)
for i in range(1,Nd):
    m.obj = pyo.Objective(expr = m.S[i]*cost_Storage, sense=pyo.minimize)


#begin = time.time()

#solve
opt = SolverFactory('ipopt', executable='C:\\ipopt.exe')
#opt = SolverFactory('couenne', executable='C:\\My_Files\\_H2_HYDROGEN_THE_GREEN\\ThisPC_MODELLING\\PYTHON-MODEL\\ENERGY SYSTEM OPTIMISATION\RHODE\\Couenne-0.3.2-win32-msvc9\\Couenne-0.3.2-win32-msvc9\\bin\\couenne.exe')

#opt.options['tol']=1e-3
#opt.solve(m)
m.results = opt.solve(m)

#detalT = time.time() - begin

#print
m.pprint()

print('\n\nStorage:',pyo.value(m.S))


print('\n\nOF Storage cost €:',pyo.value(m.obj))